Part 7: Factorization of the Resolvent Polynomial

We know from our experience of solving cubic and quartic equations that the process of solving the polynomial involves extending the field $K$ with an appropriate entity $r$ to enable factorizing $g(x)$ into a polynomial $h(x)$ of a smaller degree, such that $h(x)$ has $t$ as a root and is irreducible in $K(r)$. Here is an example: For the cubic equation, we start with the resolvent polynomial which is of degree $6 = 3!$ with (say $t=t_1$):

\[t_1 = r_1 + \omega r_2 + \omega^2 r_3\] \[t_2 = r_3 + \omega r_1 + \omega^2 r_2\] \[t_3 = r_2 + \omega r_3 + \omega^2 r_1\] \[t_4 = r_1 + \omega r_3 + \omega^2 r_2\] \[t_5 = r_2 + \omega r_1 + \omega^2 r_3\] \[t_6 = r_3 + \omega r_2 + \omega^2 r_1\]

The irreducible resolvent polynomial is:

\[g(x) = \prod_{i=1}^{6}(x - t_i)\]

After we are able to extend the field with:

\[\begin{align*} t_1^3 = (r_1 + \omega r_2 + \omega^2 r_3)^3 &= r_1^3 + r_2^3 + r_3^3 + 6r_1 r_2 r_3 \\& +3\omega(r_1^2r_2+r_2^2r_3+r_3^2r_1)\\ &+3\omega^2(r_1r_2^2+r_2r_3^2+r_3r_1^2)\\ t_4^3 = (r_1 + \omega^2 r_2 + \omega r_3)^3 &= r_1^3 + r_2^3 + r_3^3 + 6r_1 r_2 r_3 \\& +3\omega^2(r_1^2r_2+r_2^2r_3+r_3^2r_1)\\ &+3\omega(r_1r_2^2+r_2r_3^2+r_3r_1^2) \end{align*}\]

We can show that the field extension containing $t_1^3$ and $t_4^3$ can be expressed as $K(r)$, where $r = (r_1+\omega r_2+\omega^2 r_3)^3 - (r_1+\omega^2 r_2+\omega r_3)^3$ and that $r^2 \in K$.

In the new extended field, we are able to factorize $g(x)$ into the two irreducible factors $h(x) = h_1(x) = (x^3-t_1^3)$ and $h_2(x) = (x^3-t_4^3)$. The Galois Group corresponding to $g(x)$ is $S_3$, the entire permutation group of all $3! = 6$ permutations. For the resolvent polynomial $h(x)$ we have the roots:

\[t_1 = r_1 + \omega r_2 + \omega^2 r_3\] \[t_2 = r_3 + \omega r_1 + \omega^2 r_2\] \[t_3 = r_2 + \omega r_3 + \omega^2 r_1\]

The Galois Group corresponding to $h(x)$ contains $3$ permutations: The permutation of $(r_1,r_2,r_3)$ into one of ${(r_1,r_2,r_3),(r_3,r_1,r_2),(r_2,r_3,r_1)}$. These permutations correspond to $t_1 \to t_1$, $t_1 \to t_2$, and $t_1 \to t_3$. We observe that these same permutations map $t_4 \to t_4$, $t_4 \to t_6$, and $t_4 \to t_5$.

Subgroup

When $g(x)$ can be factorized into a smaller irreducible polynomial $h(x)$, such that $t$ is a root of $h(x)$, we can define a new Galois Group corresponding to $\text{Gal}(K(r_1,r_2,\ldots,r_n)/K(r))$. This new Galois Group maps roots of $h(x)$ to itself. These permutations can then be extended onto the entire field $K(r_1,r_2,\ldots,r_n)$ such that it leaves $K(r)$ unchanged. Let us denote:

\[G := \text{Gal}(K(r_1,r_2,\ldots,r_n)/K)\] \[H := \text{Gal}(K(r_1,r_2,\ldots,r_n)/K(r))\]

We know that $\sigma \in H$ maps the root $t$ of $h(x)$ to another root of $h(x)$. For any root $t’$ of $g(x)$, if $t’$ is not a root of $h(x)$, then $t’$ is a root of another factor of $g(x)$. Let us say that factor is $h_2(x)$. The factors of $g(x)$ are polynomials with coefficients from $K(r)$. When $\sigma$ is interpreted as an automorphism that fixes $K(r)$, we can say:

\[\sigma(h_2(x)) = h_2(\sigma(x))\]

Substituting $t’$ for $x$, we get: $\sigma(h_2(t’)) = h_2(\sigma(t’))$. This means $h_2(\sigma(t’))=0$. We have showed that $\sigma(t’)$ is also a root of $h_2(x)$.

$\sigma$ maps roots of each factor of $g$ onto other roots of the same factor. In this way, it maps all roots of $g(x)$ to other roots of $g(x)$. This proves that $H$ is a subgroup $G$.

Cosets and Size of Subgroup

$H$ is a subgroup of $G$. For any $\sigma \in G, \sigma \notin H$, consider the set of permutations

\[\sigma H = \{ \sigma h \hspace{2mm} | \hspace{2mm} h \in H \}\]

Let $\sigma(t) = t’$. $t’$ is a root of $g(x)$, but not $h(x)$. Let $t’$ be root of $h_2(x)$. $\sigma H$ consists of permutations that map $t \to t’ \to t’’$. We know that permutations in $H$ map roots of $h_2(x)$ to another root of $h_2(x)$. Thus, $t’’$ is also a root of $h_2(x)$. $\sigma H$ can then be interpreted as permutations that map $t$ to some root $t’’$ of $h_2(x)$.

If two permutations of $\sigma H$ map $t$ to the same root of $h_2(x)$, then we have $\sigma h_1 = \sigma h_2$ for some permutations $h_1, h_2 \in H$. But we can then compose $\sigma^{-1}$ to both of them to get $h_1 = h_2$ which is not possible. Thus, $\sigma H$ has just as many permutations as $H$. This also means that $h_2(x)$ has just as many roots as $h(x)$.

We can repeat this process till we have covered all the permutations in $G$, and equivalently, all the irreducible factors of $g(x)$. Each of these factors would have the same degree. The sets $\sigma H$ are called right-cosets of $H$. They all are of the same size. This shows that size of $H$ divides the size of $G$. This process of dividing a group into cosets of its subgroup can be carried for any group. In general, the size of a subgroup always divides the size of the group.

Polynomial Factorization

Suppose the field $K$ is extended using a $r$ such that $r^{p} \in K$, where $p$ is a prime number. The extended field is $K(r)$.

Let $U(x)$ be a polynomial with coefficients in $K(r)$. Then, $U$ can also be expressed as a polynomial with coefficients in $K$:

\[U(x, r) = a_m(r) x^m +a_{m-1}(r) x^{m-1} +\ldots +a_2(r)x^2 +a_1(r)x +a_0(r)\]

If $V(x, r)$ is a factor of $U(x, r)$, then $U(x, r) = V(x, r)W(x, r)$

\[U(x, r) - V(x,r)W(x,r) = b_m(r) x^m +b_{m-1}(r) x^{m-1} +\ldots +b_2(r)x^2 +b_1(r)x +b_0(r) = 0\]

$r$ is a root of $b_m(y)$, $b_{m-1}(y)$, $\ldots$, $b_2(y)$, $b_1(y)$, $b_0(y)$. The irreducible polynomial of $r$ over $K$ is $y^p - r^p = 0$. Thus, $y^p - r^p$ divides all of these polynomials. This means:

\[U(x, r) = V(x,r)W(x,r)\] \[U(x, \alpha r) = V(x,\alpha r)W(x,\alpha r)\] \[U(x, \alpha^2 r) = V(x,\alpha^2 r)W(x,\alpha^2 r)\] \[\vdots\] \[U(x, \alpha^{p-2} r) = V(x,\alpha^{p-2} r)W(x,\alpha^{p-2} r)\] \[U(x, \alpha^{p-1} r) = V(x,\alpha^{p-1} r)W(x,\alpha^{p-1} r)\]

where $\alpha$ is a primitive $p^{th}$ root of unity.

What we have shown is that if $V(x, r)$ divides $U(x, r)$, then $V(x, \alpha^i r)$ divides $U(x, \alpha^i r)$.

Factorization of Resolvent Polynomial

If $U(x, r) = g(x)$, then $U(x,r)$ is a polynomial with coefficients in $K$. If $h(x, r)$ is a factor of $U(x, r) = g(x)$, then all of $h(x, \alpha^i r)$ are factors of $U(x, \alpha^i r) = g(x)$. Thus, we can find $p$ different factors of $g$ from a single factor $h(x, r)$:

\[g(x) = h(x,r)q(x,r)\] \[g(x) = h(x,\alpha r)q(x,\alpha r)\] \[g(x) = h(x,\alpha^2 r)q(x,\alpha^2 r)\] \[\vdots\] \[g(x) = h(x,\alpha^{p-1} r)q(x,\alpha^{p-1} r)\]

Multiplying these, we would get:

\[g(x)^p = H(x,r)Q(x,r)\]

where:

\[H(x, r) = h(x, r)h(x, \alpha r)h(x, \alpha^2 r) \ldots h(x, \alpha^{p-1} r)\] \[Q(x, r) = q(x, r)q(x, \alpha r)q(x, \alpha^2 r) \ldots q(x, \alpha^{p-1} r)\]

The coefficients of $H(x,r)$ and $Q(x, r)$ are symmetric in $(r,\alpha r, \alpha^2 r, \ldots, \alpha^{p-1} r)$. This implies that that coefficients are from $K$. This would imply:

\[g(x)^p = H(x)Q(x)\] \[g(x)^j = H(x) = h(x, r)h(x, \alpha r)h(x, \alpha^2 r) \ldots h(x, \alpha^{p-1} r)\]

Comparing the degrees, we get:

\[\text{deg}(g) \hspace{2mm} | \hspace{2mm} p \hspace{2mm} \text{deg}(h)\]

But we also know from the previous discussion on cosets that:

\[\text{deg}(h) \hspace{2mm} | \hspace{2mm} \text{deg}(g)\]

For both these constraints to be satisfied, $\text{deg}(g)$ has to be a multiple of $\text{deg}(h)$. If $\text{deg}(g) = m\text{deg}(h)$, then $m\text{deg}(h) \hspace{2mm} | \hspace{2mm} p\text{deg}(h)$, implies that $m = 1$ or $m=p$.

$m=1$ implies, $\text{deg}(g) = \text{deg}(h)$. Or in other words, $g(x) = h(x)$. This is not possible, since we know that the field extension enables us to factorize $g$ into smaller degree factors like $h$.

$m=p$ implies, $\text{deg}(g) = p\text{deg}(h)$

\[g(x)^j = H(x) = h(x, r)h(x, \alpha r)h(x, \alpha^2 r) \ldots h(x, \alpha^{p-1} r)\]

Comparing the degrees again, $j\text{deg}(g) = p\text{deg}(h)$ gives $j=1$. Or in other words

\[g(x) = H(x) = h(x, r)h(x, \alpha r)h(x, \alpha^2 r) \ldots h(x, \alpha^{p-1} r)\]

Thus the factorization of $g$ enabled by the field extension $K(r)$ has a special form where all the factors are related to each other:

\[g(x) = h(x, r)h(x, \alpha r)h(x, \alpha^2 r) \ldots h(x, \alpha^{p-1} r)\]

Normal Subgroup

Let $t’$ be a root of $h(x,r)$. We know that $t’$ can be written as a polynomial in $t$ with coefficients in $K$: $t’ = \tau(t)$. Since $h(\tau(t),r) = 0$, we know that $t$ is a root of $h(\tau(x),r)$. Thus $h(x,r)$ is a factor of $h(\tau(x),r)$.

$h(x,r)$ being a factor of $h(\tau(x),r)$, implies that, $h(x,\alpha^i r)$ is a factor of $h(\tau(x),\alpha^i r)$. Thus, if $u$ is a root of $h(x,\alpha^i r)$, so is $\tau(u)$.

The mapping $t \to \tau(t)$ can be represented using a permutation in $H$. The permutation that maps $t \to u$ (say for $\sigma \in G$) also maps $\tau(t) \to \tau(u)$, since the coefficients of $\tau$ are from $K$. Thus, the mapping $t \to \tau(t) \to \tau(u)$ can be seen as a permutation in $H$ followed by some appropriate permutation in $G$. Thus, the permutation from $t$ to the roots of each factor can be represented as left-cosets $H\sigma$.

We already know that these same permutations can be presented as right-cosets $\sigma H$. It is a rare occurence that right-cosets are also left-cosets:

\[\sigma H = H\sigma\]

This relation holds true for all $\sigma$. In other words for any $\sigma \in G$:

\[\sigma^{-1} H\sigma =H\]

Whenever, a subgroup follows this property, we call it a Normal Subgroup.

Examples of Normal Subgroup

Normal subgroups are a rare thing. For example in the permutation group of $S_3$, there are subgroups formed by swapping two elements:

\[H = \{(r_1,r_2,r_3),(r_1,r_3,r_2)\}\]

These are not normal subgroups. The only normal subgroup turns out to be the cyclic permutation:

\[H = \{(r_1,r_2,r_3),(r_2,r_3,r_1),(r_3,r_1,r_2)\}\]

What do we know so far:

The field $K$ is extended to $K(r)$ using some $r$ such that $r^p \in K$ for some prime $p$.
The resolvent polynomial is factorized in $K(r)$. If this is possible, it can be factorized into $p$ factors: $g(x) = h(x, r)h(x, \alpha r)h(x, \alpha^2 r) \ldots h(x, \alpha^{p-1} r)$.
If $h(x)$ is the factor containing $t$ as a root, the new galois group correpsonding to $h(x)$ is a normal subgroup of the galois group corresponding to $g(x)$.

Continue Reading: Part 8: Solvability

Rediscovering Galois Theory: Part-7

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